Hi d= density of ideal gas= 0.357g/L T(Temperature) = 32.5 degree Celcius T (in

Question

Hi

d= density of ideal gas= 0.357g/L

T(Temperature) = 32.5 degree Celcius

T (in Kelvin)= 273.15 + T (in degree Celcius)

T (K)= 273.15+ 32.5 = 305.65 K

P (Pressure)= 425 mmHg

P (in atm)= P(in mmHg)/760 = 425/760 = 0.5592 atm

R= (universal gas constant)= 0.082 atm L/ (moles K)

we know that, PV= nRT

we know that, n= m/MW

where, m= mass, MW= Molecular weight

and m/V= d= density,

so, PV= (m/MW)RT

MW= (m/V)RT/P

MW= dRT/P

putting all values,

MW= 0.357*0.082*305.65/0.5592 g/gmole

MW= 16.00 g/gmole (Answer)

Thank you

Explanation / Answer

Ifan skal has a densaty of 0.357 BL gas at 32 5 and a pressaare of 425 mmHg, the molar mass of the gas is (a) 128 g/mole (b) 16.0 g/mole (c) 8.25 g/imole (d) 474 g/mole (e) none of the above Given the data he AH for the reaction A7/2F3E+4D is 3B +4D > A +SF Ar-576 k/mole )-408 kJ/mole (b) -324 kJ/mole (c) 828 kJ/mole (d) 492 kJ/mole )none of the above mass of a cube having a density of 3.25 g/cm is 58.60 g. What is the length of a side of the cube in mm? (b) 7.86 mm 18.0 mm none of the above (c) 26.2 mm (d) 2.62 mm pecific heat capacity ofcopper is 0380 J/g If the initial temperature of 75.0 g of coppers 25.0°C,wh al temperature, in °C, of the copper after absorbing 1.50 x 10 Joules of heat? (b) 25.8 (c) 30.3 (d) 11.3°C of the above is correct

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